To solve the differential equation $xy''' - xy' + 2y = 0$, we'll use a method that is often applied to such equations, assuming a solution of the form $y = x^r$. This is because the given equation is homogeneous and its coefficients are functions of $x$.

1. Assume a solution of the form $y = x^r$: $y = x^r$ Then, $y' = r x^{r-1}$ $y'' = r(r-1) x^{r-2}$ $y''' = r(r-1)(r-2) x^{r-3}$

2. Substitute these into the differential equation $xy''' - xy' + 2y = 0$: $x \left( r(r-1)(r-2) x^{r-3} \right) - x \left( r x^{r-1} \right) + 2 \left( x^r \right) = 0$

3. Simplify each term: $x \cdot r(r-1)(r-2) x^{r-3} = r(r-1)(r-2) x^{r-2}$ $x \cdot r x^{r-1} = r x^r$ $2 x^r = 2 x^r$

4. Substitute back into the equation: $r(r-1)(r-2) x^{r-2} - r x^r + 2 x^r = 0$

5. Combine like terms: $r(r-1)(r-2) x^{r-2} - r x^r + 2 x^r = 0$ Notice that the powers of $x$ must be the same for all terms to combine. Rewrite $r x^r$ and $2 x^r$ in terms of $x^{r-2}$: $r(r-1)(r-2) x^{r-2} - r x^r + 2 x^r = 0$ $r(r-1)(r-2) x^{r-2} - r x^r + 2 x^r = 0$

   Since $x^{r-2}$ and $x^r$ are different powers of $x$, equate coefficients to zero separately. Therefore, we only need to equate the polynomial coefficients to zero:

6. Form and solve the characteristic equation: $r(r-1)(r-2) = 0$

   The roots are: $r = 0, \quad r = 1, \quad r = 2$

7. Construct the general solution: Since we have three distinct roots, the general solution to the differential equation is: $y(x) = C_1 x^0 + C_2 x^1 + C_3 x^2$ $y(x) = C_1 + C_2 x + C_3 x^2$ where $C_1$, $C_2$, and $C_3$ are arbitrary constants.

Thus, the general solution to the differential equation $xy''' - xy' + 2y = 0$ is: $y(x) = C_1 + C_2 x + C_3 x^2$

Would you like more details or have any questions? Here are some related questions you might consider:

1. What is the method used to solve this differential equation called?
2. How would you solve a similar differential equation with different coefficients?
3. Can you provide an example of a non-homogeneous differential equation?
4. How does the solution change if one of the roots of the characteristic equation is repeated?
5. What if all roots of the characteristic equation are the same?
6. Can you verify the solution by substituting it back into the original differential equation?
7. What are some applications of differential equations in real-world problems?
8. How would the approach change if the differential equation had variable coefficients?

Tip: When solving differential equations, always verify your solution by substituting it back into the original equation to check for consistency.